Kind of inspired by this question, but something I've been thinking about for a while.

How much does the air in a bike tire weigh? Is it an appreciable amount? Is there a point where using a wider tire, like at 28c at 80 psi would be lighter than a 25c tire at 100psi? Obviously, this depends on the specific tires used. I don't have a scale precise enough to measure, and I don't have the math/physics knowledge to figure this out.

I seriously doubt that it's an appreciable amount, but this is a fascinating question. I hope someone has the knowledge and/or equipment to come up with an answer.
–
jimirings♦Apr 19 '14 at 14:12

Once you've got your head around the eloquent and informative answers below, there's also the schools of thought that say you should inflate tyres from bottled gases rather than air. Not only will this affect the mass of the wheel it will also affect how quickly the tyre deflates. Or possibly not.
–
PeteHApr 19 '14 at 20:38

5 Answers
5

The ideal gas law (which is a good approximation in this case) says PV=nRT where P is pressure, V is volume, n is mols of gas, R is the ideal gas law constant, and T is temperature in Kelvin.

Thus, solving for n, we see n = (PV)/(RT). Then, assuming air is made up of {gas1, gas2,...} with fractions {p1,p2,...} (so p1+p2+...=1) and corresponding molar masses {m1,m2,...}, the mass of air in a tire is (PV/(RT))(p1*m1+p2*m2+...). So, what we see is that the mass of air in a tire is directly proportional to the volume of the tire and directly proportional to the pressure in the tire, and inversely proportional to the temperature of the air in the tire.

We will make the following (reasonable) assumptions: Assume the temperature is around room temperature (293 Kelvin) and the volume of the tire regardless of pressure is the same (primarily determined by the shape of the rubber, assuming not severely under/over inflated). For convenience, air is about {nitrogen, oxygen} with {p1,p2}= {0.8,0.2} and molar masses {28 g/mol,32 g/mol}. Thus, under these assumptions (V is fixed, and T is fixed), the mass of the air in tire grows linearly with pressure.

So, the mass of air in a tire of volume V and pressure P and temperature T is about (PV/RT)(0.8*28+0.2*32) grams. It may be better to write it as "P ((V/(RT)) (0.8*28+0.2*32)) grams" noting that V/(RT) is a constant for us.

Since I don't want to put the units into wolfram alpha carefully, you can put in the entry "(7 bar* 10 gallons)/(ideal gas constant*293 Kelvin)*(0.8*28+0.2*32)" and read the result off in grams (ignoring the unit it says there) to get an estimate for the weight of air in a 7 bar (~100 psi), 10 gallon volume tire as around 313 grams. Is 10 gallons reasonable? No.

Lets be crude about estimating the volume of a tube using a torus. The volume of a torus is V=(pi*r^2)(2*pi*R) where R is the major radius and r is the minor radius. Google will calculate it for you (and has a picture of what major and minor radius is).

I can't be bothered to actually go outside and measure these things, but lets be crude and use a massive tire. Say the minor radius is 2 inches, and the major radius is 15 inches (this is probably larger the size of the tire on something like a Surly Moonlander). This has a volume of about 5 gallons. If you were a nutcase and running this at 7 bar, it would be around 150 grams of air. At a more reasonable 1 bar or 2 bar, youd be at 45 or 90 grams.

What about a thin road bike tire? Lets also assume the major radius is around 15 inches, and the minor radius is around a half inch. Thats around 0.3 gallons of volume. Plugging into our formula, at 7 bar, we see that this is about 9 grams. At 10 bar, a whopping 13.5 grams.

For reference, according to the bag of potato chips i have next to me (Lays Wavy Hickory Barbacue), one chip is about 2 grams. So, if you're a road biker and you're worried about the weight of air in your tires, note that one serving of potato chips (28 grams) is more than the air in both of your tires. All of which are much lower than even a light tire (lightest I can find is 130 grams).
–
BatmanApr 19 '14 at 16:22

+up for the examples calculated, very interesting.
–
olee22Jul 12 '14 at 8:20

What are these gallons and inches of which you speak?
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andy256Jul 13 '14 at 5:31

Just in case you prefer general knowledge over physics: the density of air at a reasonable temperature is around 1.2 kgm-3.

The volume of your tire (accepting Tom77's answer) is 0.000959m3.

So the mass of air in it at 15°C and atmospheric pressure is around 1.1g.

Then we do need one bit of physics, the relationship between mass and pressure for a given gas in a given volume and temperature is linear. This comes from just Boyle's law provided we're prepared to believe that twice as much gas at the same temperature and pressure has twice the mass. Which is a lot like saying that two buckets of water weighs twice as much as one bucket of water, so hopefully not controversial ;-) So I've cleverly(?) avoided needing to know the ideal gas law and the value of the universal gas constant in favour of a direct crib off Wikipedia measurement of air.

Atmospheric pressure is 15 psi (ish), so when you measure 80psi that's really 95, so it's 95/15 = 6.3 times as dense as the external air. So the answer is 6.3 * 1.1.

7g (0.2 ounce), at the 15°C stated by the Wikipedia article for my estimate of the density of air.

If you change the temperature from there then the pressure changes linearly, according to the combined gas law (or "Gay-Lussac's law" apparently is the name for this component of it, I had to look this up), provided you measure temperature in Kelvins not Celsius. 0°C is 273.15K. So to consider variations in temperature and pressure starting from my value, just multiply the 7g in proportion. Adding 3°C is about 1%, so the difference is smaller than my margins of error. Adding 20psi to the pressure is about 20%, or another 1g. The mass of air is already far smaller than the weight of the wheels. So pressure has more effect than temperature for the examples you give but no, it does not appreciably affect the weight of the wheels.

There's also another small confounding factor, which is that inner tubes are stretchy and so the volume does increase a bit as the pressure changes, requiring a little bit more gas. But not much.

@DanielRHicks: A value to withing 10% is often useful. You didn't specify the conditions in your original question, so the responders had to guess. In fact, nobody has specified whether the pressures are absolute or gauge (relative to atmosphere). This makes more than 10% difference at usual bike tire pressures. Just the fact that we are talking single digits of grams says the air mass difference will be dominated by tire/rim mass difference-a useful fact.
–
Ross MillikanApr 20 '14 at 17:00

@RossMillikan - I was being facetious. A value within a factor of 10 is good enough here.
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Daniel R HicksApr 20 '14 at 18:26

No one has really addressed the size versus pressure part of the question.

Nominally different sized tires will have about the same mass of air. As the size of the tire goes up the design pressure goes down. The contact patch must support the weight of the rider. Assume bike with rider is 100 lbs on the rear wheel. At 100 psi the size of the contact patch is 1 square inch. On a bigger tire you can drop the pressure down to get a bigger contact patch. At 80 psi the same rider would have a contact patch of 1.25 square inches. You cannot just reduce the pressure on a small tire to get a bigger contact patch without without banging the rim.

Lets assume the n in PV=nRT the same in all diameter tires. If so what would the relationship of diameter to pressure be? S for small and B for big

If Pb / Ps = (rS/rB)^2 then the two tires will have the same mass of air.
If the pressure is inversely proportional to diameter squared the two tires have the same mass of air.

So let's test at 25mm 100psi and see what pressure at 28mm is the same weight
Pb = (25/28)^2 * 100
Pb = 79.7 PSI

So in your example of 28c at 80 psi versus 25c tire at 100psi
The answer is almost exactly the same mass

Not the question but if you assume the same mass how does contact patch size scale with diameter. Contact patch is load / pressure
So that ratio is
(Lb / Pb) / (Ls / Ps)
but Lb = Ls so
Ps / Pb
sub in for Pb from above
Ps / Ps * (rS/rB)^2
1 / (rS/rB)^2
(rB/rS)^2

So if you keep the mass in the tire constant then the contact patch goes up with the square of the diameter. And that makes sense since area is proportional to diameter square.

Why would you keep the mass the same? Because it makes sense. Consider the force the beads must withstand. If the mass is the same then the total force on beads is the same. Same number of molecules will produce the same force. The Force is proportional to pressure * area. Force is proportional to r^2 * P.
Consider the ratio of the force on beads from big diameter to small at constant air mass.
Fb / Fs
Pb * rB^2 / Ps * rS^2
sub for Ps again with constant mass assumption
Ps * (rS/rB)^2 * rB^2 / (Ps * rS^2)
1
If you keep the number of molecules constant then the total force on the beads is constant regardless of tire diameter.

I know a lot of you are going to think I am full of BS. But various sized diameters have about the same number of molecules in them. As the diameter goes up the contact patch size goes up with the square of the diameter. So a 2" tire will nominally have 1/2 the pressure and 4 x the contact size of a 1".

Even at the lower pressure a larger diameter is less susceptible to pinch flats because it has further to travel to the rim and it builds area faster relative to deflection. I know even more of you are not going to believe me on this but even at the lower pressure the pinch resistance is proportional to the diameter squared.

Interesting question, and I enjoyed the technical answers given, but...
I am continuously amazed at the overemphasis placed on the weight of bikes. Yes it is important, but relative to other factors it is not that significant. Lets compare a 20# bike to a 24# bike. If your budget is $1000 for a new bike, would you choose a 20# bike with very good components and a so-so feel/fit, or a 24# bike with the correct feel/fit and superior components. The heavier bike is 20% heavier...but lets ask more questions. 1) On your typical ride, how many long steep climbs are there? 2) How often do you need to do long/hard accelerations? I ask these because the only time weight is a significant factor is while climbing or accelerating. in flat or nearly flat sections of a ride weight is virtually insignificant. And, BTW for all the climbs you do there is presumably equal descents, during which time more weight works to your advantage. 3) What (other than weight) can quickly sap your energy? Rough gear changes due to less effective derailleurs (even small jolts can break your rhythm and stroke) Poor body position due to improper fit. Poor technique. ( My 80 yr old mother regularly swims 1-1.5 miles, and can only do so because of a lifetime of swimming distances which require excellent breathing and stroke techniques. i can bike 40 miles NP but can only swim a few laps, because my technique sucks.) HOW MUCH DO YOU WEIGH??? If you weigh 165# then the difference between the 2 bikes, including rider weight, is 2%! Would you trade that for better components? If your buying a new bike and are more than 10# overweight, which is the vast majority of us, you can expect to save more in weight by riding more often than you can buying a lighter bike. Lastly, unless you are competing, an additional 30 seconds on your ride time is not nearly as important as how much you enjoy the ride, which comes from (among other things) overall bike quality and fit, not weight. Spend your money on overall bike feel/fit and quality of components. Save weight by riding more. Do no worry unnecessarily about 1/4 MPH or 30 seconds during a 1 hr ride. We are riding to enjoy life and the experience.